1. ## problem regarding volume

The example starts as so,

a crate has a square base of width (and length) x metres and is of height h metres. The haulage company will only transport crates provided that the sum of the length, width and height of a crate is not more than 5 metres. The firm wishes to maximise the volume V of the crate.

I want to explain why h = 5 - 2X if we want to maximise V,

and also to show that the quantity to be maximised is V = 5X^2 - 2X^3

not really sure ive got a grip on this one any help appreciated
- bobby

2. Originally Posted by bobbymellor
The example starts as so,

a crate has a square base of width (and length) x metres and is of height h metres. The haulage company will only transport crates provided that the sum of the length, width and height of a crate is not more than 5 metres. The firm wishes to maximise the volume V of the crate.

I want to explain why h = 5 - 2X if we want to maximise V,

and also to show that the quantity to be maximised is V = 5X^2 - 2X^3

not really sure ive got a grip on this one any help appreciated
- bobby
ok, so you're on the right track
now what you want to do is find the max value for x
so you do this by setting the dervative of V = 5x^2 -2x^3 to 0
then you get V'= 10x-6x^2 = 0
solve that and you get x=5/3m, or x is approx. 1.67m
and you can show that this is the max and no the min, by taking the second derivative, and since that answer will always be negative, therefore the graph in concave down, meaning its the mac value.